How to derive Schrödinger’s equation

There is a particularly neat way to derive Schrödinger’s equation, and to justify the “canonical substitution” rules for replacing energy and momentum with corresponding operators when we “quantize” an equation.

Take a particle in a potential. Its energy is given by

$$E=\frac{{\bf p}^2}{2m}+V({\bf x}),$$

or

$$E-\frac{{\bf p}^2}{2m}-V({\bf x})=0.$$

Now multiply both sides this equation by the formula $e^{i({\bf p}\cdot{\bf x}-Et)/\hbar}$. We note that this exponential expression cannot ever be zero if the part in the exponent that’s in parentheses is real:

$$\left[E-\frac{{\bf p}^2}{2m}-V({\bf x})\right]e^{i({\bf p}\cdot{\bf x}-Et)/\hbar}=0.$$

So far so good. But now note that

$$Ee^{i({\bf p}\cdot{\bf x}-Et)/\hbar}=i\hbar\frac{\partial}{\partial t}e^{i({\bf p}\cdot{\bf x}-Et)/\hbar},$$

and similarly,

$${\bf p}^2e^{i({\bf p}\cdot{\bf x}-Et)/\hbar}=-\hbar^2{\boldsymbol\nabla}e^{i({\bf p}\cdot{\bf x}-Et)/\hbar}.$$

This allows us to rewrite the previous equation as

$$\left[i\hbar\frac{\partial}{\partial t}+\hbar^2\frac{{\boldsymbol\nabla}^2}{2m}-V({\bf x})\right]e^{i({\bf p}\cdot{\bf x}-Et)/\hbar}=0.$$

Or, writing $\Psi=e^{i({\bf p}\cdot{\bf x}-Et)/\hbar}$ and rearranging:

$$i\hbar\frac{\partial}{\partial t}\Psi=-\hbar^2\frac{{\boldsymbol\nabla}^2}{2m}\Psi+V({\bf x})\Psi,$$

which is the good old Schrödinger equation.

The method works for an arbitrary, generic Hamiltonian, too. Given

$$H({\bf p})=E,$$

we can write

$$\left[E-H({\bf p})\right]e^{i({\bf p}\cdot{\bf x}-Et)/\hbar}=0,$$

which is equivalent to

$$\left[i\hbar\frac{\partial}{\partial t}-H(-i\hbar{\boldsymbol\nabla})\right]\Psi=0.$$

So if this equation is identically satisfied for a classical system with Hamiltonian $H$, what’s the big deal about quantum mechanics? Well… a classical system satisfies $E-H({\bf p})=0$, where $E$ and ${\bf p}$ are eigenvalues of the differential operators $i\hbar\partial/\partial t$ and $-i\hbar{\boldsymbol\nabla}$, respectively. Schrödinger’s equation, on the other hand, remains valid in the general case, not just for the eigenvalues.